Question

PLEASE HELP MEOWT!!!
Rewrite sin^(4)xtan^(2)x in terms of the first power of cosine.

Answer 1

Explanation:

`{ \sin(x) }^(4) { \tan(x) }^(2)`

`{ \sin(x) }^(4) \frac{ { \sin(x) }^(2) }{ { \cos(x) }^(2) }`

`\frac{ ({ {1 - \cos(x) }^(2) })^(3) }{ { \cos(x) }^(2) }`

hopefully this helps, I'm rusty with my trig identities

Answer 2

sin⁴(x)tan²(x) = (10 -15cos(2x) +6cos(4x) -cos(6x))/(16(1 +cos(2x))

Explanation:

The relevant identities are ...

`\sin^4{x}=(3-4cos((2x))+cos((4x)))/(8)\\\\\tan^2{x}=(1-cos((2x)))/(1+cos((2x)))\\\\cos((a))cos((b))=(cos((a+b))+cos((a-b)))/(2)`

Then your product is ...

`\sin^4{(x)}\tan^2{(x)}=(3-4cos((2x))+cos((4x)))/(8)\cdot(1-cos((2x)))/(1+cos((2x)))\\\\=\frac{3-4cos((2x))+cos((4x))-3cos((2x))+4\cos^2{(2x)}-cos((4x))cos((2x))}{8(1+cos((2x)))}`

Collecting terms and using the identity for the product of cosines, we get ...

`=(3-7cos((2x))+cos((4x))+4(1+cos((4x)))/(2)-(cos((6x))+cos((2x)))/(2))/(8(1+cos((2x))))\\\\=(10-15cos((2x))+6cos((4x))-cos((6x)))/(16(1+cos((2x))))`